Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}9x-3y &= -6 \\ -7x+6y &= 7\end{align*}$
Begin by moving the $x$ -term in the second equation to the right side of the equation. $6y = 7x+7$ Divide both sides by $6$ to isolate $y$ $y = {\dfrac{7}{6}x + \dfrac{7}{6}}$ Substitute this expression for $y$ in the first equation. $9x-3({\dfrac{7}{6}x + \dfrac{7}{6}}) = -6$ $9x - \dfrac{7}{2}x - \dfrac{7}{2} = -6$ Simplify by combining terms, then solve for $x$ $\dfrac{11}{2}x - \dfrac{7}{2} = -6$ $\dfrac{11}{2}x = -\dfrac{5}{2}$ $x = -\dfrac{5}{11}$ Substitute $-\dfrac{5}{11}$ for $x$ back into the top equation. $9( -\dfrac{5}{11})-3y = -6$ $-\dfrac{45}{11}-3y = -6$ $-3y = -\dfrac{21}{11}$ $y = \dfrac{7}{11}$ The solution is $\enspace x = -\dfrac{5}{11}, \enspace y = \dfrac{7}{11}$.